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This question is somewhat related to this post.

Let us consider we have a white noise current source $i_n(t)$, with a variance $\sigma_i^2$, and mean, $\mu_n=0$.

Assume that this current is passed through a system, which integrates for a finite time duration ($T_{int}$) on a capacitor $C$. One can then write an equation for the power of the noise voltage (integration of current produces voltage $V=\frac{1}{C}\int_{0}^{T}i(t)dt $ ) produced from this integration process ($\sigma_v^2$):

$$\begin{align}\sigma_{v}^{2} &=E\bigg(\frac{1}{C}\int_{0}^{T_{int}}i_{n}(t_{1})dt_{1}\frac{1}{C}\int_{0}^{T_{int}}i_{n}(t_{2})dt_{2}\bigg) \\&=\frac{1}{C^{2}}\int_{0}^{T_{int}}\int_{0}^{T_{int}}E(i_{n}(t_{1})i_{n}(t_{2}))dt_{1}dt_{2} \\&=\frac{1}{C^{2}}\int_{0}^{T_{int}}\int_{0}^{T_{int}}\sigma_i^2\delta(t_{2}-t_{1})dt_{1}dt_{2} \\&=\frac{1}{C^{2}}\int_{0}^{T_{int}}\sigma_i^2dt_{2} =\frac{1}{C^{2}}\sigma_i^2\cdot T_{int}\end{align}$$

Here $\delta$ denotes the dirac delta impulse function.

This is a similar result obtained to what was obtained in the post linked above. The problem with the above equation is that it is dimensionally inconsistent (there should be a $T_{int}^2$ factor in the numerator to make it consistent.)

Can someone kindly point out the error in the above argument?

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The problem comes from the dimension of $\sigma^2_i$, which you assume to be $[I^2]$. That's, however, not the case if you define

$$E\{i(t_1)i(t_2)\}=\sigma^2_i\delta(t_2-t_1)\tag{1}$$

Note that the dimension of a Dirac delta impulse is the inverse of the dimension of its argument, which is $[1/T]$ or $[Hz]$. So you get for the dimension of $\sigma^2_i$ $[I^2T]$ or $[I^2/Hz]$. You can also see this by looking at the power spectral density (the Fourier transform of $(1)$), which is simply $\sigma^2_i$, and the dimension of which must be $[I^2/Hz]$. With this dimension of $\sigma^2_i$, the dimension of your result is correct.

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  • $\begingroup$ Thank you, for your answer. I wasn't aware of the dimension of the Dirac delta being inverse of the argument. That clears it up. $\endgroup$ – Saqib Shah Jan 15 '18 at 14:54

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